02157 Functional Programming Finite Trees (I)

Transcription

1 Finite Trees (I) nsen 1 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

2 Overview Finite Trees Algebraic Datatypes. Non-recursive type declarations: Disjoint union (Lecture 4) Recursive type declarations: Finite trees Recursions following the structure of trees Illustrative examples: Search trees Expression trees File systems... Mutual recursion, layered pattern, polymorphic type declarations 2 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

3 Finite trees A finite tree is a value which may contain a subcomponent of the same type. Example: A binary search tree Br Br 9 Br Br 7 Lf Br 21 Br Lf 2 Lf Lf 13 Lf Lf 25 Lf Condition: for every node containing the value x: every value in the left subtree is smaller then x, and every value in the right subtree is greater than x. 3 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

4 Example: Binary Trees A recursive datatype is used to represent values which are trees. type Tree = Lf Br of Tree*int*Tree;; Lf;; val it : Tree = Lf Br;; val it : Tree * int * Tree -> Tree = <fun:clo@4> The two parts in the declaration are rules for generating trees: Lf is a tree if t 1, t 2 are trees, n is an integer, then Br(t 1, n, t 2 ) is a tree. The tree from the previous slide is denoted by: Br(Br(Br(Lf,2,Lf),7,Lf), 9, Br(Br(Lf,13,Lf),21,Br(Lf,25,Lf))) 4 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

5 Binary search trees: Insertion Recursion on the structure of trees Constructors Lf and Br are used in patterns The search tree condition is an invariant for insert let rec insert i = function Lf -> Br(Lf,i,Lf) Br(t1,j,t2) as tr -> match compare i j with 0 -> tr n when n<0 -> Br(insert i t1, j, t2) _ -> Br(t1,j, insert i t2);; val insert : int -> Tree -> Tree Example: let t1 = Br(Lf, 3, Br(Lf, 5, Lf));; let t2 = insert 4 t1;; val t2 : Tree = Br (Lf,3,Br (Br (Lf,4,Lf),5,Lf)) 5 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

6 Binary search trees: member and inorder traversal let rec memberof i = function Lf -> false Br(t1,j,t2) -> match compare i j with 0 -> true n when n<0 -> memberof i t1 _ -> memberof i t2;; val memberof : int -> Tree -> bool In-order traversal let rec inorder = function Lf -> [] Br(t1,j,t2) -> inorder inorder t2;; val tolist : Tree -> int list gives a sorted list inorder(br(br(lf,1,lf), 3, Br(Br(Lf,4,Lf), 5, Lf)));; val it : int list = [1; 3; 4; 5] 6 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

7 Deletions in search trees Delete minimal element in a search tree: Tree -> int * Tree let rec delmin = function Br(Lf,i,t2) -> (i,t2) Br(t1,i,t2) -> let (m,t1 ) = delmin t1 (m, Br(t1,i,t2));; Delete element in a search tree: int -> Tree -> Tree let rec delete j = function Lf -> Lf Br(t1,i,t2) -> match compare i j with n when n<0 -> Br(t1,i,delete j t2) n when n>0 -> Br(delete j t1,i,t2) _ -> match t2 with Lf -> t1 _ -> let (m,t2 ) = delmin t2 Br(t1,m,t2 );; 7 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

8 Parameterize type declarations The programs on search trees just requires an ordering on elements they no not need to be integers. A polymorphic tree type is declared as follows: type Tree< a> = Lf Br of Tree< a> * a * Tree< a>;; Program texts are unchanged (though polymorphic now), for example let rec insert i = function... Br(t1,j,t2) as tr -> match compare i j with... ;; val insert: a -> Tree< a> -> Tree< a> when a: comparison let ti = insert 4 (Br(Lf, 3, Br(Lf, 5, Lf)));; val ti : Tree<int> = Br (Lf,3,Br (Br (Lf,4,Lf),5,Lf)) let ts = insert "4" (Br(Lf, "3", Br(Lf, "5", Lf)));; val ts : Tree<string> = Br (Lf,"3",Br (Br (Lf,"4",Lf),"5",Lf)) 8 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

9 Higher-order functions for tree traversals For example satisfies let rec infoldback f t e = match t with Lf -> e Br(t1,x,t2) -> let er = infoldback f t2 e infoldback f t1 (f x er);; val infoldback: ( a -> b -> b) -> Tree< a> -> b -> b infoldback f t e = List.foldBack f (inorder t) e It traverses the tree without building the list- For example: let ta = Br(Br(Br(Lf,-3,Lf),0,Br(Lf,2,Lf)),5,Br(Lf,7,Lf));; inorder ta;; val it : int list = [-3; 0; 2; 5; 7] infoldback (-) ta 0;; val it : int = 1 9 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

10 Example: Expression Trees type Fexpr = Const of float X Add of Fexpr * Fexpr Sub of Fexpr * Fexpr Mul of Fexpr * Fexpr Div of Fexpr * Fexpr;; Defines 6 constructors: Const: float -> Fexpr X : Fexpr Add: Fexpr * Fexpr -> Fexpr Sub: Fexpr * Fexpr -> Fexpr Mul: Fexpr * Fexpr -> Fexpr Div: Fexpr * Fexpr -> Fexpr 10 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

11 Symbolic Differentiation D: Fexpr -> Fexpr A classic example in functional programming: let rec D = function Const _ -> Const 0.0 X -> Const 1.0 Add(fe1,fe2) -> Add(D fe1,d fe2) Sub(fe1,fe2) -> Sub(D fe1,d fe2) Mul(fe1,fe2) -> Add(Mul(D fe1,fe2),mul(fe1,d fe2)) Div(fe1,fe2) -> Div( Sub(Mul(D fe1,fe2),mul(fe1,d fe2)), Mul(fe2,fe2));; Notice the direct correspondence with the rules of differentiation. Can be tried out directly, as tree are just values, for example: D(Add(Mul(Const 3.0, X), Mul(X, X)));; val it : Fexpr = Add (Add (Mul (Const 0.0,X),Mul (Const 3.0,Const 1.0)), Add (Mul (Const 1.0,X),Mul (X,Const 1.0))) 11 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

12 Expressions: Computation of values Given a value (a float) for X, then every expression denote a float. compute : float -> Fexpr -> float let rec compute x = function Const r -> r X -> x Add(fe1,fe2) -> compute x fe1 + compute x fe2 Sub(fe1,fe2) -> compute x fe1 - compute x fe2 Mul(fe1,fe2) -> compute x fe1 * compute x fe2 Div(fe1,fe2) -> compute x fe1 / compute x fe2;; Example: compute 4.0 (Mul(X, Add(Const 2.0, X)));; val it : float = DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

13 Mutual recursion. Example: File system d 1 a 1 d 2 a 4 d 3 a 2 d 3 a 5 a 3 A file system is a list of elements an element is a file or a directory, which is a named file system 13 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

14 Mutually recursive type declarations are combined using and type FileSys = Element list and Element = File of string Dir of string * FileSys let d1 = Dir("d1",[File "a1"; Dir("d2", [File "a2"; Dir("d3", [File "a3"])]); File "a4"; Dir("d3", [File "a5"]) ]) The type of d1 is? 14 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

15 Mutually recursive function declarations are combined using and Example: extract the names occurring in file systems and elements. let rec namesfilesys = function [] -> [] e::es -> (nameselement (namesfilesys es) and nameselement = function File s -> [s] Dir(s,fs) -> s :: (namesfilesys fs) ;; val namesfilesys : Element list -> string list val nameselement : Element -> string list nameselement d1 ;; val it : string list = ["d1"; "a1"; "d2"; "a2"; "d3"; "a3"; "a4"; "d3"; "a5"] 15 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012

16 Summary Finite Trees concepts illustrative examples Notice the strength of having trees as values. Notice that polymorphic types and mutual recursion are NOT biased to trees. 16 DTU Informatics, Technical University of Denmark Finite Trees (I) MRH 11/10/2012